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This is a mixed physics and conceptual exercise: In my story, we are aboard a multi-generational mission to resettle Humanity on a new planet in our galaxy.

In the internal life of the ship inhabitants, I want to say it is less relevant to use Earth's years to measure time, and I assume that with the ongoing generation cycles, a new measurement of time will take hold. In my specific story, I assume it will be the Light-Year position of the ship in the journey (i.e "I was born on LY38, got married on LY47.2" and so on)

NOW - as the ship changes its velocity along its trajectory, each LY will take different "Earth Years" to travel (putting aside for just one moment relativistic effects).

The ship starts from near-Earth, accelerating at 0.8g until it reaches 0.3c. It then "flips over" so that the rear propulsion engines are now pushing back to decelerate from 0.3c at a lighter push of 0.6g until full stop (there are narrative reasons for the 0.8/0.6 figures so let's assume those are the numbers).

Could you assist me with the math of: a) how long would the full journey take? b) how many light-years will we travel in total? c) if at 0.3c a LY is approx. 40 Earth months, does the duration linearly extend as the speed decreases (i.e what would a LY "cost" in months at 20%, 40%, 60%, 80% of the journey?)

And most importantly (this is the conceptual part) - as readers of fiction, would you rather "let it slide" when a story uses familiar bits like counting people's age in years, or would you make the effort to learn the new vocabulary and mentally understand that "this occured on LY28" means about 20 years earlier than the present?

  • much appreciated!
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    $\begingroup$ I am sure the math has been explained already somewhere here, and asked in different flavors several times already. Have done any search before posting? $\endgroup$
    – L.Dutch ♦
    May 9, 2020 at 15:42
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    $\begingroup$ As a reader of fiction, I require the author to adhere to Chekhov's Gun. If the exact dates aren't important, don't make me learn them. Heinlein's Time For The Stars (1956) dealt with relativistic travel by walking the reader through it over the course of the book in terms of generational differences - the young protagonist returns home after a journey and begins dating his similarly-aged grand-niece (ewww)...but never mentions the difference in terms of years or dates. $\endgroup$
    – user535733
    May 9, 2020 at 16:08
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    $\begingroup$ (1) A light-year is a measure of distance, not of time. (2) 0.3 c is 299,792,458 / 3 = 99 930 819.3 meters/second, and g is 9.80665 m/s². As we all learned in 4th grade, the ship will accelerate for (299,792,458 / 3) / (9.80665 × 0.8) = 12,737,634.6 seconds (about 147 days and 11 hours), and will then decelerate for 16,983,512.8 seconds (about 196 days and 13 hours). In total the ship will travel for 344 days, less than one year, covering a distance of 1,485,029,305,604.4 km, or 0.157 light-years. $\endgroup$
    – AlexP
    May 9, 2020 at 16:09
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    $\begingroup$ The only thing that really matters is ship board time. I don't think that mixing distances (light years) and time duration (years) is useful. $\endgroup$
    – Slarty
    May 9, 2020 at 16:10
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    $\begingroup$ As a reader, measuring time elapsed in terms of distance is a bit odd. It would be like saying that you slept for 120 miles on a road trip; having to convert from distance to time takes you out of the story a bit. At least, that's my personal opinion- others may find it adds to the immersion of the story. $\endgroup$
    – Jack
    May 9, 2020 at 17:00

3 Answers 3

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The inhabitants of the ship would know where they are going and know the day and year length of the planet they are going to because of advanced astronomical observations before the ship left Earth.

The most logical approach would be to immediately adopt a clock and calendar compatible with their new home and measure everything against that. The trajectory and velocity of the ship should be known well in advance so it should even be possible to align the calendar with the seasons on arrival. Relativistic effects could then be ignored as only ship board time would be relevant.

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309 days. 0.126 light years.

AlexP already answered in the comments, as is his wont. I am ok with that. I suspect 4th grade math in Romania was more advanced than 4th grade math in South Dakota.

Also I made a jpeg!

https://www.calculatorsoup.com/calculators/physics/velocity_a_t.php

maths!

In this calculator v is final velocity, u is starting velocity, a is acceleration and t is time it take to go from speed u to speed v.

89937737 is 0.3c

7.84 m/s^2 is 0.8g and 5.88 m/s^2 is 0.6g.

Acceleration leg takes 11471700s = 132 days. Deceleration leg takes 15295500 or 177 days.
That is 309 days. Instead of light years you could measure your trip in school years! haw haw haw snort.

How far? 11471700 s of smoothly increasing velocity = 132 days at the average of all velocities traveled which is 89937737 + 0 / 2 or 44968868.5 m/s.
44968868.5 * 11471700 = 5.1586937e+14 or 0.054 light years. 515869370000 kilometers though!

Now decreasing velocity. 15295500 * 44968868.5 (same average velocity) = 6.8782133e+14 m or 687821330000 km or 0.0727 light years.

Total trip is 0.126 light years. Out past Pluto for sure, but nothing much around. The nearest star to the sun is Alpha Centauri at 4.2 light years.

hmm. My math did not come out the same as AlexP. At least it is within an order of magnitude. Persons who spot errors please correct in the comments.

At the end of the day you want a nice long trip for your story. The solution is to coast at 0.3 c for a while and not start slowing down right away. Given that the acceleration and deceleration times are pretty trivial compared to the time at 0.3 to cross interstellar space, you can just consider your 0.3c cruising speed as your ships speed for the whole trip.

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    $\begingroup$ Alex P did c/3 rather than c*0.3- I imagine that is where the difference lies. $\endgroup$
    – Jack
    May 9, 2020 at 16:56
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    $\begingroup$ OK maybe not 4th grade but definitely in 5th grade we were introduced to mysteries such as speed equals acceleration times time and distance equals average speed times time. I clearly remember problems of the form a train leaves from city A towards city B with speed so-and-so; so much time later, another train leaves from city B towards city A with another speed so-and-so; at what time and at what point will the two trains cross? It appears that they are pan-European, given that the French even have a dedicated name for them: problème de trains (related to the problème de robinets). $\endgroup$
    – AlexP
    May 9, 2020 at 23:35
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    $\begingroup$ You forgot to add that 'adjustment for relativity' fudge factor in the denominator, as proposed by Einstein, that changes the Newtonian equation into a relativistic equation, adjusting for speeds approaching c, and time/space dilation. You know, that (c-v) factor that becomes 'divide by zero error' at the speed of light? The one they do not teach in junior school? Or probably even senior high school in South Dakota? $\endgroup$
    – Justin Thyme the Second
    May 10, 2020 at 16:59
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    $\begingroup$ @JustinThymetheSecond - Ignorance is curable and I am ready for the cure! Go ahead and addend the question and show how to use that fudge factor. Or just throw it in a comment if you prefer. $\endgroup$
    – Willk
    May 11, 2020 at 0:42
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    $\begingroup$ To add two velocities: Newton: it is just V(t) = V(1) + V(2), a simple addition; Einstein: V(t) = (V(1)+V(2)) / (1 + (V(1)*V(2)/c^2)) The change is the denominator, which effectively limits the addition of the velocities so they are never more than c, the fudge factor that makes the statement 'nothing can travel faster than c' appear to be true, and that produces a divide-by-zero error when you subtract velocities in the opposite direction, denominator is (1+(-1) = 0) when v(1) = c and v(2) = -c (in the opposite direction). $\endgroup$
    – Justin Thyme the Second
    May 11, 2020 at 4:16
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I posted this as a comment, but then I realized it was also an answer to part of your question implied in the last paragraph, so here it is.

'(putting aside for just one moment relativistic effects)' I put my generation ship in a modified Alcubierre bubble, even though it was not going FTL and I did not need it as a drive, just to eliminate the relativistic effects on living things and hand-wave them away. We really do not know how life reacts to relativistic effects, i.e. circadian rhythms. Do they follow the rules of time dilation? What about the gestation period for living things? Do biological processes accelerate/slow down with increased speed? Does life have a separate 'space/time dimension'than physical things?

So once in this bubble, it would be 'life as usual' no matter how 'fast' the ship was going. What I can not attest to, is the effect on 'artificial gravity' that doing so would result in. Would inhabitants in a bubble actually feel and respond to artificial gravity due to acceleration of the bubble? So I spun my ship anyway to produce artificial gravity (it is an asteroid so it had a huge mass to begin with), and I gave it a constant velocity once it reached my coasting speed. You might have to hand wave that detail away as well, since we do not really know the answer. Giving something a constant acceleration means it is no longer a continuous 'relativistic frame of reference'.

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